Chemical Engineering - Mass Transfer - Discussion

Both C ad D correct.

As pressure increases dew point increases as pressure decreases dew point decreases.

Compression and expansion of air affect its dew point. Generally speaking, compression increases dew point, and expansion (i.e De-compression) lowers dew point. For example, consider compressed air leaving a dryer at 200 PSIG with a pressure dew point of -40° F/C at 200 psig. If the pressure is eventually reduced to 100 psig, the pressure dew point will fall to -50 °F at 100 psig. If the air is further expanded to 5 psig, the pressure dew point becomes -77° F at 5 psig.

At dew point vapour convert in liquid by using latent heat at constant temp according to latent haet definition.

Assume we have pot in this pot some vapours are present if pressure increase in this pot vapour will be condensed but in another condition, we have liquid in pot at constant volume let 2 atm pressure on it if we decrease pressure then the B P will decrease in liquid is flashed and more vapours will be formed so and vapour molecules loss it's energy and get cooled and after that condensed and d p will be increased.

For a given temperature D.P. is independent of Pressure. But when both temperature and pressure increases D.P. also increases and relative humidity decreases accordingly.

Liquids have high pressure than gasses. So, at high-pressure vapor condensate at high temperature. Less cooling is required for condensation at high pressure.

Dew point is independent of temperature and directly proportional of pressure so C and D both option is correct.

Both c & d were correct answer because dew point is directly proportional to the pressure as simple as that.

Since, the dew point is directly proportional to the pressure, so both the option C&D is correct.

When vapor pressure increase D.P also increase or not? Please someone give description.

D. P. Proportional pressure.

So, both C&D is correct.

Yeah, Both should be right. I also agree @Alpesh.